The area of the garden is \(10 \times 15 = 150\) square meters. The walkway adds 2 meters to each side of the garden. So, the new length is \(15 + 2 + 2 = 19\) meters, and the new width is \(10 + 2 + 2 = 14\) meters. The total area of the garden with the walkway is \(19 \times 14 = 266\) square meters. To find the area of the walkway, subtract the garden's area from the total area: \(266 - 150 = 116\) square meters.

Let the width of the park be \(w\) meters. The length is \(w+10\) meters. The perimeter is given by \(P=2(L+W)\), so \(120 = 2((w+10)+w)\). Simplifying gives \(120 = 2(2w+10)\), which becomes \(60 = 2w+10\). Subtracting 10 gives \(50 = 2w\), so the width \(w=25\) meters. The length is \(25+10=35\) meters. The area of the park is \(L \times W = 35 \times 25 = 875\) square meters.

The amount of fencing required is the perimeter of the garden. The area of the square is given as 225 square feet. The formula for the area of a square is \(A = s^2\), where \(s\) is the side length. So, \(s^2 = 225\). To find the side length, take the square root of the area: \(s = \sqrt{225} = 15\) feet. The perimeter of a square is \(P = 4s\). So, the required fencing is \(4 \times 15 = 60\) feet.

First, find the side length of the square. If the area is 64 cm², the side length is \(\sqrt{64} = 8\) cm. The perimeter of the square is \(4 \times 8 = 32\) cm. The rectangle has the same perimeter, 32 cm. The formula for the perimeter of a rectangle is \(P = 2(L+W)\). We have \(32 = 2(10+W)\). Dividing by 2 gives \(16 = 10+W\), so the width \(W = 6\) cm. The area of the rectangle is \(L \times W = 10 \times 6 = 60\) square centimeters.

Let the width be \(w\). The length is \(2w\). The area is \(L \times W\), so \((2w)(w) = 98\). This gives \(2w^2 = 98\). Dividing by 2, we get \(w^2 = 49\), so the width \(w = 7\) feet. The length is \(2 \times 7 = 14\) feet. The perimeter is \(2(L+W) = 2(14+7) = 2(21) = 42\) feet.

Let the side length of Square Y be \(s_y\). Its perimeter is \(4s_y\). Let the side length of Square X be \(s_x\). Its perimeter is \(4s_x\). We are given \(4s_x = 4(4s_y)\), which simplifies to \(4s_x = 16s_y\), and further to \(s_x = 4s_y\). The area of Square Y is \(A_y = (s_y)^2\). The area of Square X is \(A_x = (s_x)^2 = (4s_y)^2 = 16(s_y)^2\). Therefore, the area of Square X is 16 times the area of Square Y.

Let the original length and width be L and W. The original perimeter is \(P = 2(L+W)\) and the original area is \(A = LW\). The new length and width are 3L and 3W. The new perimeter is \(P_{new} = 2(3L+3W) = 2(3(L+W)) = 3(2(L+W)) = 3P\). The new area is \(A_{new} = (3L)(3W) = 9(LW) = 9A\). So, the perimeter is tripled and the area becomes nine times as large.

The length of the wire is the perimeter of both shapes. For the square, the perimeter is 60 inches, so each side is \(60 \div 4 = 15\) inches. The area of the square is \(15 \times 15 = 225\) square inches. For the rectangle, the perimeter is also 60 inches. Using \(P=2(L+W)\), we have \(60 = 2(18+W)\). This simplifies to \(30 = 18+W\), so the width is \(12\) inches. The area of the rectangle is \(18 \times 12 = 216\) square inches. The difference in areas is \(225 - 216 = 9\) square inches.

The perimeter of the original rectangle is \(2(15+20) = 2(35) = 70\) inches. When a square is cut from a corner, two segments of the original perimeter are removed, but two new segments of the same length are added. For example, at one corner, a 4-inch segment from the length and a 4-inch segment from the width are conceptually removed. They are replaced by the two inner sides of the cut-out square, each 4 inches long. The net change to the perimeter at each corner is \(-4 - 4 + 4 + 4 = 0\). Since this happens at all four corners, the perimeter of the new shape is the same as the original perimeter, which is 70 inches.

The formula for the perimeter of a rectangle is \(P = 2(L+W)\). Substitute the expressions for length and width: \(52 = 2((3x+2) + x)\). Simplify the expression inside the parentheses: \(52 = 2(4x+2)\). Distribute the 2: \(52 = 8x + 4\). Subtract 4 from both sides: \(48 = 8x\). Divide by 8 to find \(x=6\). Now find the dimensions: the width is \(x = 6\) meters, and the length is \(3(6) + 2 = 18 + 2 = 20\) meters. The area is length times width: \(A = 20 \times 6 = 120\) square meters.